Benefit-Timing Pitfalls in Complement Identities

Why Benefit Timing Matters

A life insurance value and an annuity value can be complements because their payments partition the possible outcomes over time. But the identity depends on when each payment occurs. A whole-life insurance benefit payable at the end of the year of death pairs with an annuity-due value, whose payments occur at the beginnings of years. Substituting an annuity-immediate value without adjusting the identity creates a timing error.

The distinction is only one payment date, but that date matters: an annuity-due payment is made at time 0 if the insured is alive, while the first annuity-immediate payment is made at time 1 if the insured survives to then. The discount factor applied to those payments is therefore different.

Definition: A discrete whole-life insurance with unit benefit payable at the end of the year of death has present-value random variable \(Z=v^{K+1}\), where \(K\) is the curtate future lifetime. Its expected present value is \(A_x\). A whole-life annuity-due with unit payments at times \(0,1,2,\ldots\), while the insured is alive, has expected present value \(\ddot{a}_x\). The corresponding whole-life annuity-immediate has payments at times \(1,2,3,\ldots\) while the insured is then alive, and value \(a_x\).

The Complement Identity and Its Timing

Let \(i\) be the effective annual interest rate, \(v=1/(1+i)\), and \(d=1-v=i/(1+i)\). Under the standard discrete model, the whole-life insurance benefit is paid at the end of the year of death and the annuity-due pays at the beginning of each year of life. Then

A_x+d\ddot{a}_x=1, \qquad\text{so}\qquad A_x=1-d\ddot{a}_x.

The annuity-immediate value satisfies \(\ddot{a}_x=1+a_x\): the due annuity has the time-0 payment, worth 1, plus the later payments of the immediate annuity. Therefore the equivalent identity using \(a_x\) is

$$ A_x=1-d(1+a_x). $$

It is not \(A_x=1-da_x\). That expression omits the time-0 payment embedded in the due annuity. It may look like a minor adjustment, but it changes the insurance value by exactly \(d\).

Why the Whole-Life Identity Holds

The identity can be justified by examining each possible year of death. Suppose death occurs in year \(k+1\), so the insured is alive at times \(0,1,\ldots,k\). On that path, the insurance pays \(v^{k+1}\). The due annuity pays at times \(0\) through \(k\), with present value \(\sum_{j=0}^{k}v^j\). Multiplying that annuity value by \(d\) gives

d\sum_{j=0}^{k}v^j =d\frac{1-v^{k+1}}{1-v} =1-v^{k+1},

because \(d=1-v\). Adding the insurance present value gives \(v^{k+1}+(1-v^{k+1})=1\) for every possible year of death. Taking expectations over the mortality model preserves this equality, giving \(A_x+d\ddot{a}_x=1\). This argument requires the stated payment pattern: changing the payment dates changes the pathwise present values.

Identify the insurance timing.
For \(A_x\), the unit benefit is paid at the end of the year of death.

Identify the annuity timing.
The matching complement uses payments at the beginnings of years of life, so it uses \(\ddot{a}_x\).

Apply the identity.
Calculate \(A_x=1-d\ddot{a}_x\), or convert \(a_x\) to \(\ddot{a}_x=1+a_x\) first.

Check the result.
Verify that the insurance and annuity values use the same interest and mortality basis and the specified payment dates.

Worked Example: Correcting an Immediate-Annuity Substitution

A hypothetical whole-life contract has a unit death benefit payable at the end of the year of death. At an effective annual interest rate of 5%, the annuity-immediate value on the contract’s mortality basis is \(a_x=7.5\). Find \(A_x\), and calculate the result of the common incorrect substitution \(1-da_x\). The mortality assumption is the annual independent survival model with constant survival probability \(p=63/68\) and death probability \(q=5/68\).

Method. Convert the supplied immediate-annuity value to a due value, then use the complement identity. This avoids pairing the year-end death benefit with the wrong annuity timing.

First, \(v=1/1.05=20/21\), so \(d=1-v=1/21\). The annuity-due value is

\ddot{a}_x=1+a_x=1+7.5=8.5.

Thus

A_x=1-\frac{1}{21}(8.5) =1-\frac{17}{42} =\frac{25}{42} \approx 0.5952.

For a check from the stated mortality model, the one-year discounted survival factor is \(vp=(20/21)(63/68)=15/17\). The geometric annuity-immediate value is \(vp/(1-vp)=(15/17)/(2/17)=7.5\), as given. The insurance value is \(vq/(1-vp)=(20/21)(5/68)/(2/17)=25/42\), confirming the result.

The incorrect substitution gives

1-da_x =1-\frac{1}{21}(7.5) =1-\frac{15}{42} =\frac{27}{42} \approx 0.6429.

The incorrect value exceeds the correct value by \(2/42=1/21=d\), because it omits the time-0 payment when converting from \(a_x\) to \(\ddot{a}_x\).

Finite-Term Complements: Keep the Survival Boundary

For a term insurance with unit benefit payable at the end of the year of death, define \(A_{x:\overline{n}|}\) as its expected present value over the first \(n\) years. Define \(\ddot{a}_{x:\overline{n}|}\) as the temporary annuity-due value with payments at times \(0,\ldots,n-1\), provided the insured is alive at each payment time. Write \({}_np_x\) for the probability of surviving \(n\) years. The finite-term complement includes a survival-contingent boundary value:

A_{x:\overline{n}|} +d\ddot{a}_{x:\overline{n}|} +v^n{}_np_x =1.

The final term is the present value, in expectation, of 1 paid at time \(n\) if the insured survives to that time. Without it, the complement would leave out the outcome in which the insured survives the entire term. The identity is based on annual discrete payments and a death benefit at the end of the year of death; a different benefit timing requires a matching derivation rather than automatic reuse of this formula.

Worked Example: A Term Insurance Value With the Boundary Included

A hypothetical three-year term insurance pays 1 at the end of the year of death if death occurs within the term. The annual effective interest rate is 4%. Assume an annual independent survival model with \(p=0.8\) and \(q=0.2\) in each year. Find the term insurance value using the complement identity, and compare it with a direct calculation.

Method. Calculate the temporary annuity-due and the discounted probability of surviving to year 3, then use the finite-term identity. The key trap is omitting the survival boundary term.

Here \(v=25/26\), \(d=1/26\), and the one-year discounted survival factor is \(vp=(25/26)(4/5)=10/13\). Therefore

\ddot{a}_{x:\overline{3}|} =1+\frac{10}{13}+\left(\frac{10}{13}\right)^2 =\frac{399}{169}.

The discounted survival probability to year 3 is

v^3{}_3p_x =\left(\frac{10}{13}\right)^3 =\frac{1000}{2197}.

The complement identity now gives

A_{x:\overline{3}|} =1-\frac{1}{26}\frac{399}{169}-\frac{1000}{2197} =\frac{4394-399-2000}{4394} =\frac{1995}{4394} \approx 0.4540.

For a direct check, death in years 1, 2, and 3 has probabilities \(0.2\), \(0.8(0.2)=0.16\), and \(0.8^2(0.2)=0.128\). Thus

A_{x:\overline{3}|} =0.2v+0.16v^2+0.128v^3 =\frac{5}{26}\left(1+\frac{10}{13}+\frac{100}{169}\right) =\frac{1995}{4394}.

The direct and complement calculations agree. If the survival term were omitted, the result would be too large by \(1000/2197\), the discounted probability-weighted value of surviving to the end of the term.

Endowment Values and a Second Timing Check

If the contract also pays 1 at time \(n\) when the insured survives \(n\) years, its expected present value includes the pure endowment value \(v^n{}_np_x\). Adding that value to the term insurance identity cancels the survival boundary term:

A^{1}_{x:\overline{n}|} =A_{x:\overline{n}|}+v^n{}_np_x =1-d\ddot{a}_{x:\overline{n}|}.

Here \(A^{1}_{x:\overline{n}|}\) denotes the endowment insurance value: the death benefit is payable at the end of the year of death during the term, and the survival benefit is payable at time \(n\). The identity still pairs that benefit pattern with the temporary annuity-due, not the temporary annuity-immediate.

Worked Example: An Endowment Value and the Wrong Convention

A hypothetical two-year endowment insurance pays 1 at the end of the year of death if death occurs within two years, or 1 at time 2 if the insured survives. The effective annual interest rate is 6%. Assume annual independent survival with \(p=0.9\) and \(q=0.1\). Calculate the endowment value, then show what happens if the temporary annuity-immediate is incorrectly substituted into the due identity.

Method. Use the endowment complement with the temporary annuity-due, then verify by adding the death and survival present values. The common mistake is using the immediate annuity without accounting for its different payment dates.

Here \(v=50/53\), \(d=3/53\), and \(vp=(50/53)(9/10)=45/53\). The two-payment annuity-due value is

\ddot{a}_{x:\overline{2}|} =1+\frac{45}{53} =\frac{98}{53}.

Therefore

A^{1}_{x:\overline{2}|} =1-\frac{3}{53}\frac{98}{53} =\frac{2809-294}{2809} =\frac{2515}{2809} \approx 0.8950.

Check directly. The first-year death benefit has value \(vq=(50/53)(1/10)=265/2809\). The second-year death benefit has value \(v^2pq=(45/53)(5/53)=225/2809\). The survival benefit has value \(v^2p^2=(45/53)^2=2025/2809\). Their sum is \((265+225+2025)/2809=2515/2809\), as required.

The corresponding temporary annuity-immediate value includes payments at times 1 and 2:

a_{x:\overline{2}|} =\frac{45}{53}+\left(\frac{45}{53}\right)^2 =\frac{4410}{2809}.

Incorrectly using \(1-da_{x:\overline{2}|}\) gives

1-\frac{3}{53}\frac{4410}{2809} =\frac{135647}{148877} \approx 0.9111,

which is not the endowment value \(2515/2809\approx0.8950\). The annuity-immediate has a payment at time 2, whereas the temporary annuity-due used in the identity pays at times 0 and 1. The payment dates cannot be interchanged just because both annuities have two potential payments.

Common Exam Traps

The complement identity is a compact calculation, but only after the contract’s payment pattern has been identified. Before substituting values, check the insurance benefit date, the annuity payment dates, and whether survival to a fixed endpoint produces another benefit or boundary value.

Using \(a_x\) where \(\ddot{a}_x\) is required. Convert with \(\ddot{a}_x=1+a_x\). For whole life, \(A_x=1-d(1+a_x)\), not \(1-da_x\).
Dropping the finite-term survival value. For term insurance, retain \(v^n{}_np_x\) in the complement. It represents survival through the term, not a death claim.
Assuming all annuities with the same number of payments are interchangeable. Payment dates determine discounting and survival conditions; state them explicitly.
Mixing bases. Insurance and annuity values in an identity must use the same interest rate, mortality assumption, and contract period.

A reliable check is to write a short payment timeline before calculating. If the benefit is payable at the end of the year of death, match it to the due annuity in the stated identity; if the annuity value supplied is immediate, convert it before substitution.

Check Your Understanding

For each problem, identify the payment timing before using a complement identity.

A whole-life policy pays 1 at the end of the year of death. At an effective annual interest rate of 4%, \(a_x=9\). Find \(A_x\), then calculate the incorrect value from \(1-da_x\). (Answer: \(A_x=8/13\approx0.6154\); incorrect value \(17/26\approx0.6538\).)
A whole-life annuity-due value is \(\ddot{a}_x=12\), and the effective annual interest rate is 5%. The associated insurance pays 1 at the end of the year of death. Find \(A_x\). (Answer: \(A_x=9/21=3/7\approx0.4286\).)
A two-year term insurance pays 1 at the end of the year of death. The annual effective interest rate is 5%, and survival is independent by year with \(p=0.8\), \(q=0.2\). Find its expected present value using the finite-term complement. (Answer: \(148/441\approx0.3356\).)
A three-year term policy has \(v^3{}_3p_x=0.30\), \(\ddot{a}_{x:\overline{3}|}=2.8\), and \(i=5\%\). Its unit death benefit is payable at the end of the year of death. Find the term insurance value. (Answer: \(1-0.30-(1/21)(2.8)=0.5667\).)
A whole-life contract has \(a_x=8.5\) at an effective annual interest rate of 4%, with a unit death benefit payable at the end of the year of death. Find the correct insurance value and the value from the incorrect immediate-annuity substitution. (Answer: Correct \(A_x=33/52\approx0.6346\); incorrect value \(35/52\approx0.6731\).)

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